An arch is in the form of a semi-ellipse. It | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The arch is a semi-ellipse with a total width of $8 \, m$ and a maximum height of $2 \, m$ at the centre.This implies the length of the semi-major

Vedclass · Accepted Answer

$1.56$

Vedclass · Answer

(A) The arch is a semi-ellipse with a total width of $8 \, m$ and a maximum height of $2 \, m$ at the centre.This implies the length of the semi-major axis $a = 4 \, m$ and the semi-minor axis $b = 2 \, m$.The equation of the ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Substituting the values,we get $\frac{x^2}{16} + \frac{y^2}{4} = 1$.We need the height at a point $1.5 \, m$ from one end. Since the total width is $8 \, m$,the end is at $x = 4$. $A$ point $1.5 \, m$ from the end corresponds to $x = 4 - 1.5 = 2.5$.Substituting $x = 2.5$ into the equation:$\frac{(2.5)^2}{16} + \frac{y^2}{4} = 1$$\frac{6.25}{16} + \frac{y^2}{4} = 1$$\frac{y^2}{4} = 1 - \frac{6.25}{16} = \frac{16 - 6.25}{16} = \frac{9.75}{16}$$y^2 = 4 	imes \frac{9.75}{16} = \frac{9.75}{4} = 2.4375$$y = \sqrt{2.4375} \approx 1.56 \, m$.

An arch is in the form of a semi-ellipse. It is $8 \, m$ wide and $2 \, m$ high at the centre. Find the height of the arch at a point $1.5 \, m$ from one end. (in $, m$)

Similar Questions

An angle of intersection of the curves,$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and $x^{2}+y^{2}=ab$,where $a > b$,is :

If a circle $(x-1)^2+y^2=r^2$ touches the ellipse $x^2+4y^2=16$ internally,then $r=$

The foci of the ellipse $9x^2 + 25y^2 = 225$ are

If $a$ and $c$ are positive real numbers and the ellipse $\frac{x^2}{4c^2} + \frac{y^2}{c^2} = 1$ has four distinct points in common with the circle $x^2 + y^2 = 9a^2$,then

The ellipse having its foci $(0, \pm 1)$ and major axis of length $\sqrt{5}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module